dijkstra堆优化+链式前向星的例题

dijkstra堆优化+链式前向星的例题

一.Invitation Cards

poj - 1511 Invitation Cards

1.题意

在一个有向图中，计算从1到n中每个结点的最少费用，然后再求n个结点（2~n）到1的最少费用

2.分析

从第一个结点到各个结点的最少费用 ——> 单源最短路径
从各个结点到第一个结点的最少费用 ——> 反向建图 + 单源最短路径
所以用 反向建图 + 两次dijkstra

3.代码

#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>

using namespace std;

const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
typedef long long ll;

int t,n,r,head[maxn],head1[maxn],dis[maxn],dis2[maxn],cnt,cnt1;

struct edge{
	int to,dis,next;
}e[maxn],re[maxn];

void add(int u,int v,int d)
{
	e[++cnt].dis = d;
	e[cnt].to = v;
	e[cnt].next = head[u];
	head[u] = cnt;
}

void readd(int u,int v,int d)
{
	re[++cnt1].dis = d;
	re[cnt1].to = v;
	re[cnt1].next = head1[u];
	head1[u] = cnt1;
}

struct node{
	int index,dist;
	node(int _index=0,int _dist=0):index(_index),dist(_dist){};
	bool operator < (const node &x) const
	{
		return dist > x.dist;
	}
};

void dij()
{
	fill(vis,vis+maxn,false);
	priority_queue <node> q;
	for(int i = 1; i <= n; i++)
	{
		dis[i] = INF;
	}
	
	dis[1] = 0;
	q.push(node(1,0));
	
	while(!q.empty())
	{
		node t = q.top();
		q.pop();
		int x = t.index, d = t.dist;
		
		if(d > dis[x])		continue;
		
		for(int i = head[x]; i ; i = e[i].next)
		{
			int y = e[i].to;
			int dt = dis[x] + e[i].dis;
			if(dt < dis[y])
			{
				dis[y] = dt;
				q.push(node(y,dis[y]));
			}
		}
	}
}

void redij()
{
	fill(vis,vis+maxn,false);
	priority_queue <node> q;
	for(int i = 1; i <= n; i++)
	{
		dis2[i] = INF;
	}
	
	dis2[1] = 0;
	q.push(node(1,0));
	
	while(!q.empty())
	{
		node t = q.top();
		q.pop();
		int x = t.index, d = t.dist;

		if(d > dis2[x])		continue;

		for(int i = head1[x]; i ; i = re[i].next)
		{
			int y = re[i].to;
			int dt = dis2[x] + re[i].dis;
			if(dt < dis2[y])
			{
				dis2[y] = dt;
				q.push(node(y,dis2[y]));
			}
		}
	}
}

int main()
{
	scanf("%d",&t);
	while(t--){
		
		memset(head,0,sizeof(head));
		memset(head1,0,sizeof(head1));
		cnt = 0,cnt1 = 0;
		ll ans = 0;
		
		scanf("%d %d",&n,&r);
		for(int i = 1; i <= r; i++)
		{
			int u,v,d;
			scanf("%d%d%d",&u,&v,&d);
			add(u,v,d);
			readd(v,u,d);
		}
		
		dij();
		redij();
		
		for(int i = 1; i <= n; i++)
			ans += dis[i] + dis2[i];
		printf("%lld\n",ans);
	}
	return 0;
}

二.Roadblocks

poj - 3255 Roadblocks

1.题意

在无向图中，求第1个结点到第n个结点的次短路，一条边可以经过无数次

2.分析

要么是到第u个点是最短路，然后再加上u–>v这条边
要么是到第u个点是次短路，然后再加上u–>v这条边

所以次短路是可以是由最短路+边或者次短路+边更新，所以我们需要将最短路与次短路都进入队列，先更新最短路，然后再去更新次短路

3.代码

#include <stdio.h>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>

using namespace std;

const int maxn = 2e5 + 10;
const int N = 6000;
const int INF = 0x3f3f3f3f;
typedef long long ll;

int n,r,head[N],dis[N],dis2[N],cnt;

struct edge{
	int to,dis,next;
}e[maxn];

void add(int u,int v,int d)
{
	e[++cnt].dis = d;
	e[cnt].to = v;
	e[cnt].next = head[u];
	head[u] = cnt;
}

struct node{
	int index,dist;
	node(int _index=0,int _dist=0):index(_index),dist(_dist){}
	bool operator < (const node &x) const
	{
		return dist > x.dist;
	}
};

void dij()
{
	priority_queue <node> q;
	for(int i = 1; i <= n; i++)
	{
		dis[i] = INF;
		dis2[i] = INF;
	}
	
	dis[1] = 0;
	q.push(node(1,0));
	
	while(!q.empty())
	{
		node t = q.top();
		q.pop();
		int x = t.index, d = t.dist;
		if(d > dis2[x])		continue;
		for(int i = head[x]; i ; i = e[i].next)
		{
			int y = e[i].to;
			int dt = d + e[i].dis;
			if(dt < dis[y])
			{
				swap(dt,dis[y]);
				q.push(node(y,dis[y]));
			}
			if(dis2[y] > dt && dis[y] < dt)
			{
				dis2[y] = dt;
				q.push(node(y,dis2[y]));
			}
		}
	}
	
	printf("%d\n",dis2[n]);
}

int main()
{
	scanf("%d %d",&n,&r);
	for(int i = 1; i <= r; i++)
	{
		int u,v,d;
		scanf("%d%d%d",&u,&v,&d);
		add(u,v,d);
		add(v,u,d);
	}
	
	dij();
	
	return 0;
 }